The section ring and its filtrations

Proof. Quartic del Pezzo surfaces of types $D_4$ and $D_5$ each have a unique isomorphism class over $\mathbb C$; see [13, §3.4]. Their anticanonical equations are also displayed in [10, (1.4)–(1.5)]. Permuting variables and rescaling gives the equations above. An isomorphism of anticanonically embedded surfaces is induced by a linear change of coordinates.

Every fibre is smooth along $C$, since the two hyperplanes cutting out $C$ have independent differentials on $V$ there. Thus $p\notin C$, giving the stated form of $\ell$. Along the fibre, the first quadric has differential $du$ and the second has zero differential at $p$. Smoothness of $V$ therefore forces the coefficient of $t\ell$ in the second quadric to be nonzero. The implicit-function theorem then eliminates $u,t$ and gives the regular parameters. ◻

Proof. The complete-intersection ideal $\mathcal I_C=(\ell,t)$ has the power resolution $$0\to\mathcal O_V(-m-1)^m\to\mathcal O_V(-m)^{m+1} \to\mathcal I_C^m\to0.$$ After twisting by $2m$, intermediate cohomology vanishes because a complete intersection of two quadrics is arithmetically Cohen–Macaulay. Thus its global sections are generated by degree-$m$ polynomials multiplied by the degree-$m$ monomials in $\ell,t$.

The extended Rees presentation above has no extra $e$-torsion. Indeed, modulo $e$ the two equations define the affine cone over the complete-intersection curve $C$, so have height two. The two substituted equations form a regular sequence, and $e$ is regular on their quotient. After inverting $e$ this quotient is the coordinate ring of $V$ with $e^{\pm1}$ adjoined. It is therefore the desired extended Rees ring. Taking the indicated component gives the assertion; solving the degree equations gives the monomial bounds. ◻

Proof. All omitted terms have larger weight: the changes replacing $x_0$ by $\ell$ add $u$ times a positive-weight linear form to the first equation, and the only term in the second equation involving $t$ with minimal weight is $t\ell$. The two initial equations form a regular sequence. Lemma 5.1 consequently identifies the polynomial weight-associated graded ring with the displayed complete intersection.

We check that this is also the geometric filtration, including cancellations. The equations obtained by setting $e=0$ or $a=0$ are $$(vw,z^2+u(v+w))\quad\text{or}\quad(v^2+zw,z^2+uw),$$ both of height two. Since a complete intersection is Cohen–Macaulay, $e,a$ are non-zero-divisors in $R_0$. Localizing at $ae$ and setting $u'=u/(ea)$ and similarly for $v,z,w$, and $b'=b/a$, eliminates $u',b'$ and gives a polynomial ring in $v',z',w'$ with $a^{\pm1},e^{\pm1}$ adjoined. Thus $R_0$ is a domain. The degrees of $a,e$ form a lattice basis, so removing their common Laurent factor is injective on each fixed bidegree. This identifies its weight filtration with the local monomial valuation. In particular, a nonzero global initial form cannot disappear locally.

The same height-two argument applies after setting $b=0$. After localization the remaining equation is the irreducible $D_4$ or $D_5$ tangent polynomial in the example above. Hence $R_0/(b)$ is a domain. It follows by induction that $(b^q):(ae)^\infty=(b^q)$: reduce a putative divisibility relation modulo $b$, cancel $b$, and repeat. The same argument works for every multiplier outside $(b)$. At the generic point of the tangent curve all native coordinates are nonzero and their weights have gcd one. Its geometric multiplicity is therefore one. These facts identify the geometric curve filtration with divisibility by $b^q$.

Finally choose leading monomials $vw,z^2$ in type $D_4$, and $v^2,uw$ in type $D_5$. In each case these are relatively prime, so the two equations are a Gröbner basis. The standard monomials are as stated and form a free $\mathbb C[b]$-basis after allowing the other homogeneous variables. Thus they respect the second filtration too. ◻

Proof. Let $D_m(s)$ be the number of monomials of $\mathcal P_{(s,-m)}$, and $T_m(s)$ their total weight. If $h$ is the exponent sum in $a,b$, then $$D_m(s)=\sum_{h=m}^s(h+1)\binom{s-h+3}{3},\quad T_m(s)=\sum_{h=m}^s(h+1)\binom{s-h+3}{3} \left(\frac{w_*(s-h)}4+\frac{dh}2\right).$$ For $r=s-m\ge0$, binomial convolution gives $$\begin{aligned} D_m(s)&=m\binom{r+4}{4}+\binom{r+5}{5},\\ T_m(s)&=w_*\left(m\binom{r+4}{5}+\binom{r+5}{6}\right)\\ &\quad+d\left[\binom{m+1}{2}\binom{r+4}{4} +(m+1)\binom{r+4}{5}+\binom{r+4}{6}\right]. \end{aligned}$$ Empty sums are zero. The two degree-two relations and their Koszul syzygy give $$\begin{aligned} N_m&=D_m(2m)-2D_m(2m-2)+D_m(2m-4),\\ M_m&=T_m(2m)-2T_m(2m-2)+T_m(2m-4)\\ &\quad-(d_1+d_2)D_m(2m-2)+(d_1+d_2)D_m(2m-4). \end{aligned}$$ The last two terms account for the weights of the relations and syzygy themselves. Expansion yields $$\begin{aligned} M_m={}&\left(\frac{5w_*}6+\frac{11d}6-\frac{5(d_1+d_2)}{12}\right)m^4\\ &+\left(\frac{w_*}3+\frac{7d}3+\frac{d_1+d_2}2\right)m^3\\ &+\left(\frac{2w_*}3+\frac{13d}6-\frac{d_1+d_2}{12}\right)m^2 +\left(\frac{w_*}6+\frac{2d}3\right)m. \end{aligned}$$ Dividing by $mN_m$ gives the formula. It yields $6$ for $(w_*,d,d_1,d_2)=(11,6,4,6)$ and $8$ for $(15,8,6,8)$.

For the $b$-order, the leading monomials of both relations have $b$-exponent zero. Their deletion and inclusion–exclusion therefore give the coefficient of $d$ in the unshifted second difference of $T_m$. The total is $$K_m=\frac{11}{6}m^4+\frac73m^3+\frac{13}{6}m^2+\frac23m.$$ Lemma 20.4 makes this the actual curve-order sum. Its normalized limit is $11/16$. ◻

Proof. In the notation of Lemma 20.5, a monomial with exponent sum $h$ in $a,b$ has exponent $h-m$ in $e$. Thus the ambient total weight is $$\sum_{h=m}^s(h+1)\binom{s-h+3}{3} \left(\frac{w_*(s-h)}4+\frac{(w_a+w_b)h}2+w_e(h-m)\right).$$ Apply the same Koszul second difference. The first two coefficients were computed in Lemma 20.5. The coefficient of $w_e$ is $3/8$. One can also recover it without expansion: the two degree functions $H$ and $E$ give the constant weights $2m$ and $-m$ on the $(2m,-m)$ component. Substitution of the $E$-degrees, namely $w_a=w_b=-1$, $w_e=1$ and $d_i=0$, forces $-22/16+3/8=-1$.

For the last assertion, take the logarithmic weight derivative of the Hilbert series. The new variable and relation contribute $(w_r-d_3)z^L/(1-z^L)$ times the unweighted Hilbert series of $R$. Their normalized contribution is $$(w_r-d_3)\lim_{m\to\infty} \frac{\sum_{k=1}^m N_{m-k}}{mN_m} =\frac14(w_r-d_3).$$ Indeed, the Hilbert polynomial gives the exact identity $\sum_{k=1}^mN_{m-k}=(2m^4+m^2)/3$. The ordinary Hilbert series remains unchanged: the factor $1-z^L$ for the relation cancels that of the new variable. ◻